Is cos^4(β)-sin^4(β)=cos(2β) ?

The answer to whether cos^4(β)-sin^4(β)=cos(2β) is True

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prove cos^4(β)-sin^4(β)=cos(2β)

prove

cos^{4} (β) - sin^{4} (β) = cos (2 β)

True

Solution steps

cos^{4} (β) - sin^{4} (β) = cos (2 β)

Manipulating left side

cos^{4} (β) - sin^{4} (β)

Factor

cos^{4} (β) - sin^{4} (β) : (cos^{2} (β) + sin^{2} (β)) (cos (β) + sin (β)) (cos (β) - sin (β))

cos^{4} (β) - sin^{4} (β)

Rewrite

cos^{4} (β) - sin^{4} (β)

(cos^{2} (β))^{2} - (sin^{2} (β))^{2}

cos^{4} (β) - sin^{4} (β)

Apply exponent rule:

a^{b c} = (a^{b})^{c}

sin^{4} (β) = (sin^{2} (β))^{2}

= cos^{4} (β) - (sin^{2} (β))^{2}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

cos^{4} (β) = (cos^{2} (β))^{2}

= (cos^{2} (β))^{2} - (sin^{2} (β))^{2}

= (cos^{2} (β))^{2} - (sin^{2} (β))^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

(cos^{2} (β))^{2} - (sin^{2} (β))^{2} = (cos^{2} (β) + sin^{2} (β)) (cos^{2} (β) - sin^{2} (β))

= (cos^{2} (β) + sin^{2} (β)) (cos^{2} (β) - sin^{2} (β))

Factor

cos^{2} (β) - sin^{2} (β) : (cos (β) + sin (β)) (cos (β) - sin (β))

cos^{2} (β) - sin^{2} (β)

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

cos^{2} (β) - sin^{2} (β) = (cos (β) + sin (β)) (cos (β) - sin (β))

= (cos (β) + sin (β)) (cos (β) - sin (β))

= (cos^{2} (β) + sin^{2} (β)) (cos (β) + sin (β)) (cos (β) - sin (β))

= (cos (β) + sin (β)) (cos (β) - sin (β)) (cos^{2} (β) + sin^{2} (β))

Rewrite using trig identities

(cos (β) + sin (β)) (cos (β) - sin (β)) (cos^{2} (β) + sin^{2} (β))

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

= (cos (β) + sin (β)) (cos (β) - sin (β)) \cdot 1

Simplify

= (cos (β) + sin (β)) (cos (β) - sin (β))

Expand

(cos (β) + sin (β)) (cos (β) - sin (β)) : cos^{2} (β) - sin^{2} (β)

(cos (β) + sin (β)) (cos (β) - sin (β))

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = cos (β), b = sin (β)

= cos^{2} (β) - sin^{2} (β)

= cos^{2} (β) - sin^{2} (β)

Use the Double Angle identity:

cos^{2} (β) - sin^{2} (β) = cos (2 β)

= cos (2 β)

= cos (2 β)

We showed that the two sides could take the same form

\Rightarrow True

\frac{cot ( θ ) - 1}{1 - tan ( θ )} = cot (θ)

cot (x) = \frac{1 + cot ( x )}{1 + tan ( x )}

\frac{1}{- sin ( x )} + sin (x) = - cos (x) cot (x)

6 cos^{2} (B) - 6 sin^{2} (B) = 12 cos^{2} (B) - 6

cos^{3} (x) = \frac{3}{4} cos (x) + \frac{1}{4} cos (3 x)

Is cos^4(β)-sin^4(β)=cos(2β) ?
The answer to whether cos^4(β)-sin^4(β)=cos(2β) is True